Question: Simplify the following expression: $y = \dfrac{7x^2+5x- 2}{7x - 2}$
Explanation: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(7)}{(-2)} &=& -14 \\ {a} + {b} &=& &=& {5} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-14$ and add them together. Remember, since $-14$ is negative, one of the factors must be negative. The factors that add up to ${5}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-2}$ and ${b}$ is ${7}$ $ \begin{eqnarray} {ab} &=& ({-2})({7}) &=& -14 \\ {a} + {b} &=& {-2} + {7} &=& 5 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({7}x^2 {-2}x) + ({7}x {-2}) $ Factor out the common factors: $ x(7x - 2) + 1(7x - 2)$ Now factor out $(7x - 2)$ $ (7x - 2)(x + 1)$ The original expression can therefore be written: $ \dfrac{(7x - 2)(x + 1)}{7x - 2}$ We are dividing by $7x - 2$ , so $7x - 2 \neq 0$ Therefore, $x \neq \frac{2}{7}$ This leaves us with $x + 1; x \neq \frac{2}{7}$.